Observe that the sum of the co-effs. of odd-powered terms of g(x)=3-7=-4,g(x)=3−7=−4, & that of even-powered =2-6=-4.=2−6=−4. Hence, (x+1)(x+1) is a factor of g(x).g(x).
Now we arrange the terms of g(x)g(x) in such a way that (x+1)(x+1) can be taken out as a common factor from g(x)g(x) as shown below :-
g(x)=3x^3+2x^2-7x-6,g(x)=3x3+2x2−7x−6,
=3x^3+3x^2-x^2-x-6x-6,=3x3+3x2−x2−x−6x−6,
=3x^2(x+1)-x(x+1)-6(x+1),=3x2(x+1)−x(x+1)−6(x+1),
=(x+1)(3x^2-x-6).=(x+1)(3x2−x−6).
To work out the zeroes of the quadr. poly. 3x^2-x-6 = f(x), say,3x2−x−6=f(x),say, we compare it with the std. qudr. poly. ax^2+bx+c,ax2+bx+c, to give us,
a=3, b=-1, c=-6.a=3,b=−1,c=−6.
We will use the formula to find the zeroes of f(x)f(x).
If alpha, betaα,β are the zeroes of the std. qudr. poly., by the formula,
alpha,beta =(-b+-sqrtDelta)/(2a), where, Delta=b^2-4ac=1-4*3*(-6)=1+72=73.
Hence, alpha,beta=(1+-sqrt73)/6.
Altogether, the zeroes are -1,(1+sqrt73)/6,(1-sqrt73)/6.
Taking sqrt73~=8.544, the zeroes are -1, 1.591, -1.257.
